How many real numbers satisfy x+x^(−1) so that it is a Whole number. I came across this in a discrete mathematics book, and i wonder if my solution is correct. From looking at y=x+x^(−1) I see only 2 possibilities, x ={1;-1;} Was this just so easy or am I missing something

bolton8l 2022-10-01 Answered
How many real numbers satisfy x + x 1 so that it is a Whole number.
I came across this in a discrete mathematics book, and i wonder if my solution is correct.
From looking at y = x + x 1
I see only 2 possibilities, x ={1;-1;}
Was this just so easy or am I missing something?
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Answers (2)

domino671v
Answered 2022-10-02 Author has 8 answers
For any integer n we want
n = x + 1 x x 2 n x + 1 = 0
We need Δ = n 2 4 0 for the above equation to have solutions, so | n | 2
And the solutions are x = n ± n 2 4 2 (note that x 1 = n n 2 4 2 in that case).
For n = 2 you get the ( 1 , 1 ) solution
For n = 2 you get ( 1 , 1 ) solution
For n = 4 you get the nice ( 2 + 3 , 2 3 ) solution
and so on ...
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Aidyn Crosby
Answered 2022-10-03 Author has 3 answers
Guide:
Let
x + 1 x = k
x 2 + 1 = k x
This is a quadratic equation. For which values of k does this has a real solution?
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